Question 12.CSGP.130: Reevaluate the combined Brayton and Rankine cycles in Proble......
Reevaluate the combined Brayton and Rankine cycles in Problem 12.114. For a more realistic case assume the air compressor, the air turbine, the steam turbine and the pump all have an isentropic efficiency of 87%.
a) From Air Tables, A.7: P _{ r 1}=1.0913, \quad h _1=298.66, \quad h _5=475.84 \,kJ / kg
\begin{aligned}& s _2= s _1 \Rightarrow P _{ r 2 S }= P _{ r 1}\left( P _2 / P _1\right)=1.0913 \times 14=15.2782 \\& T _{2 S }=629 \,K , \quad h _{2 S }=634.48 \\& w _{ SC }= h _1- h _{2 S }=298.66-634.48=-335.82\, kJ / kg \\& w _{ C }= w _{ SC } / \eta_{ SC }=-335.82 / 0.87=-386= h _1- h _2 \Rightarrow h _2=684.66 \,kJ / kg\end{aligned}
At T _3=1523.2\, K : P _{ r 3}=515.493, h _3=1663.91 \,kJ / kg
\dot{ m }_{ AIR }=\dot{ Q }_{ H } /\left( h _3- h _2\right)=\frac{60000}{1663.91-684.66}= 6 1 . 2 7 ~ k g / s
b)
\begin{aligned}& P _{ r 4 S }= P _{ r 3}\left( P _4 / P _3\right)=515.493(1 / 14)=36.8209 \\& \Rightarrow T _{4 S }=791 \,K , \quad h _{4 S }=812.68 \,kJ / kg \\& w _{ ST }= h _3- h _{4 S }=1663.91-812.68=851.23\, kJ / kg \end{aligned}
w _{ T }=\eta_{ ST } \times w _{ ST }=0.87 \times 851.23=740.57= h _3- h _4 \Rightarrow h _4=923.34 \,kJ / kg
Steam cycle: – w _{ SP } \approx 0.00101(12500-10)=12.615 \,kJ / kg
\begin{aligned}& – w _{ P }=- w _{ SP } / \eta_{ SP }=12.615 / 0.85=14.84 \,kJ / kg \\& h _6= h _9- w _{ P }=191.83+14.84=206.67 \,kJ / kg\end{aligned}
At 12.5 MPa, 500 °C: h _7=3341.7 \,kJ / kg , \quad s _7=6.4617 \,kJ / kg K
\dot{ m }_{ H _2 O }=\dot{ m }_{ AIR } \frac{ h _4- h _5}{ h _7- h _6}=61.27 \frac{923.34-475.84}{3341.7-206.67}= 8 . 7 4 6 ~ k g / s
c)
\begin{gathered}s _{8 S }= s _7=6.4617=0.6492+ x _{8 S } \times 7.501, \quad x _{8 S }=0.7749 \\h _{8 S }=191.81+0.7749 \times 2392.8=2046.0 \,kJ / kg \\w _{ ST }= h _7- h _{8 S }=3341.7-2046.0=1295.7 \,kJ / kg \\w _{ T }=\eta_{ ST } \times w _{ ST }=0.87 \times 1295.7=1127.3 \,kJ / kg\end{gathered}
\begin{aligned}\dot{ W }_{ NET } & =\left[\dot{ m }\left( w _{ T }+ w _{ C }\right)\right]_{ AIR }+\left[\dot{ m }\left( w _{ T }+ w _{ P }\right)\right]_{ H _2 O } \\& =61.27(740.57-386.0)+8.746(1127.3-14.84) \\& =21725+9730=31455 \,kW =31.455\, MW \\\eta_{ TH }= & \dot{ W }_{ NET } / \dot{ Q }_{ H }=31.455 / 60= 0 . 5 2 4\end{aligned}